This Exam P sample reference tests Exponential Distribution. For an exponential lifetime with mean θ, the variance is θ² and the percentile rank of a value x is 100F(x). Evaluating the exponential CDF at x = θ² gives 100(1 − exp(−θ)), which is choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis keeps only an exponential survival term and also squares the parameter in the exponent rather than dividing the lifetime by the mean.
BThis reports a survival percentage instead of a percentile and therefore omits the leading one from the CDF.
CThis replaces the parameter-dependent standardized lifetime by the fixed value one.
DThis retains θ² in the exponent after substitution, missing the cancellation in θ²/θ.
Original practice · fully worked
Original variant: coating maintenance percentile
A protective coating has an exponentially distributed service life with mean 4 months. A maintenance team replaces the coating at 6 months if it has not already failed. The 6-month mark is approximately what percentile of the service-life distribution?
A 22.3rd
B 39.3rd
C 77.7th
D 86.5th
E 95.0th
Variant answer in brief
Standardizing 6 months by the 4-month mean gives 1.5. The exponential CDF is 1 − exp(−1.5) = 0.77687, so 6 months is about the 77.7th percentile.
Setup
Setup
Use the exponential CDF with the stated mean as its scale.
F(x)=1−e−x/4
Model
Model
A percentile rank is 100 times the cumulative probability at the specified service life.
k=100F(6)
Compute
Compute
Evaluate the standardized time and the corresponding cumulative probability.
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